Simplify the following expression: $y = \dfrac{7x^2+41x+30}{x + 5}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(7)}{(30)} &=& 210 \\ {a} + {b} &=& &=& {41} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $210$ and add them together. The factors that add up to ${41}$ will be your ${a}$ and ${b}$ When ${a}$ is ${6}$ and ${b}$ is ${35}$ $ \begin{eqnarray} {ab} &=& ({6})({35}) &=& 210 \\ {a} + {b} &=& {6} + {35} &=& 41 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({7}x^2 +{6}x) + ({35}x +{30}) $ Factor out the common factors: $ x(7x + 6) + 5(7x + 6)$ Now factor out $(7x + 6)$ $ (7x + 6)(x + 5)$ The original expression can therefore be written: $ \dfrac{(7x + 6)(x + 5)}{x + 5}$ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ This leaves us with $7x + 6; x \neq -5$.